CSI 445/660 – Network Science – Fall 2015
Solutions to Homework I
Problem 1: Construct an undirected graph without edge weights that satisfies all of the following
conditions: (i) the number of nodes in the graph is 16, (ii) each node has degree of at most 4 and
(iii) the diameter of the graph is equal to 6.
Generalize the idea behind your construction for the above problem so that for any integer n ≥ 4
such that n is a perfect square (i.e., n = k 2 for some integer k), your construction can be used to
produce an undirected graph with n nodes so that the degree of each node is at most 4 and the
√
diameter of the graph is close to 2 n.
Solution: Consider the following graph, which contains 16 nodes arranged as a 4 × 4 grid.
a
b
In this graph, the degree of each node is at most 4. Further, its diameter is 6, which is the length
of any shortest path between the two nodes marked as a and b. (Any path between a and b must
traverse at least 3 horizontal edges and 3 vertical edges.)
To generalize the above construction for any integer n = k 2 , we construct a k × k grid graph.
(The above figure is the special case with k = 4.) The diameter of the graph is determined by a
pair of opposite corner vertices; any path between such a pair of vertices must traverse at least k − 1
√
horizontal edges and k − 1 vertical edges. Thus, the diameter of the graph is 2k − 2 = 2 n − 2, since
√
k = n.
Problem 2: Consider the following graph in which edges are labeled s or w to indicate whether
they represent strong or weak ties respectively.
g
a
h
s
s
s
w
s
w
c
b
w
d
s
e
f
w
i
Add a minimum number of undirected edges to the graph so that each node satisfies the strong
triadic closure property.
Your answer must show the graph that results after you add all the necessary edges. You must
also indicate the reason for the addition of each new edge.
1
Solution: The new edges needed to ensure that each node satisfies the strong triadic closure
property are shown below as dashed lines.
g
a
h
s
w
s
w
s
d
c
b
s
w
e
f
s
w
i
The reasons for the new edges are as follows.
1. Edge {b, d} was added because of the strong ties {a, b} and {a, d}.
2. Edge {b, f } was added because of the strong ties {a, b} and {a, f }.
3. Edge {d, f } was added because of the strong ties {a, d} and {a, f }.
4. Edge {a, h} was added because of the strong ties {a, f } and {f, h}.
5. Edge {a, i} was added because of the strong ties {b, a} and {b, i}.
Problem 3: This problem has two parts. For each part, the graph example that you present must
satisfy both of the following conditions: (i) it must be connected and (ii) it must have at least 5
nodes.
Recall that a bridge in a graph is an edge whose removal disconnects the graph. Further, an
edge {x, y} of a graph is a local bridge if x and y don’t have any common neighbors.
(a) Show an example of an undirected graph G1 such that each edge of G1 is a bridge.
(b) Show an example of an undirected graph G2 such that each edge of G2 is a local bridge
but not a bridge.
Solution:
Part (a): Consider the following graph, which is a simple path with 5 nodes.
a
b
c
d
e
In the above graph, each edge is a bridge since deleting any edge splits the graph into two components.
Note: You can verify that if a graph is connected and each of its edges is a bridge, the graph must
be a tree.
Part (b): Consider the following graph, which is a simple cycle with 5 nodes.
a
b
c
d
e
In the above graph, none of the edges is a bridge since deleting any edge doesn’t split the graph into
two or more components. However, each edge is a local bridge since for any edge {x, y}, the two end
points x and y don’t have any common neighbor.
2